Calculate the number of moles of solute present in each of the following solutions: (a) \(255 \mathrm{~mL}\) of \(1.50 \mathrm{M} \mathrm{HNO}_{3}(a q),(\mathbf{b})\) \(50.0 \mathrm{mg}\) of an aqueous solution that is \(1.50 \mathrm{~m} \mathrm{NaCl}\), (c) \(75.0 \mathrm{~g}\) of an aqueous solution that is \(1.50 \%\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) by mass.

For the first part, we have been given the volume of the solution and its molarity (concentration). Molarity is defined as moles of solute per liter of solution. We can use the formula: moles = Molarity × Volume (in liters). First, we need to convert the volume from milliliters (mL) to liters (L) by dividing by 1000. \( V_{L} = \frac{255}{1000} \mathrm{L} = 0.255 \mathrm{L} \) Now, we can calculate the moles of solute: \( \mathrm{moles} = \left(1.50 \frac{\mathrm{moles}}{\mathrm{L}}\right)(0.255 \mathrm{L}) = 0.3825 \mathrm{moles} \) So, in solution (a), there are 0.3825 moles of HNO₃ present.

In the second part, we are given the mass of the solution and its molality (concentration). Molality is defined as moles of solute per kilogram of solvent. We can use the formula: moles = Molality × Mass of solvent (in kg). First, we need to convert the mass from milligrams (mg) to kilograms (kg) by multiplying by 0.001. \( m_{\mathrm{kg}} = 50.0\cdot 0.001 \mathrm{kg} = 0.05 \mathrm{kg} \) Now, we can calculate the moles of solute: \( \mathrm{moles} = \left(1.50 \frac{\mathrm{moles}}{\mathrm{kg}}\right)(0.05 \mathrm{kg}) = 0.075 \mathrm{moles} \) So, in solution (b), there are 0.075 moles of NaCl present.

For the third part, we are given the mass of the solution and the mass percent of sucrose present. Mass percent (%) is defined as the mass of solute divided by the mass of the solution, multiplied by 100. First, we need to find the mass of the solute. We can use the formula: Mass of solute = (Mass percent × Mass of solution) / 100 Mass of sucrose = (1.50% × 75.0 g) / 100 = 1.125 g To calculate the moles of sucrose present, we need to convert the mass of sucrose to moles using the molar mass of sucrose, which is: \( \mathrm{M_{Sucrose}} = 12(\mathrm{C}) + 22(\mathrm{H}) + 11(\mathrm{O}) \) \( \mathrm{M_{Sucrose}} = (12\times 12.01)+(22\times 1.01)+(11\times 16) \mathrm{g~mol}^{-1} = 342.3 \mathrm{g~mol}^{-1} \) Using the mass of sucrose and its molar mass, we can calculate the moles of sucrose: \( \mathrm{Moles} = \frac{\mathrm{Mass}}{\mathrm{Molar~mass}} \) \( \mathrm{Moles} = \frac{1.125\,\mathrm{g}}{342.3\,\mathrm{g~mol^{-1}}} = 0.00329~\mathrm{moles} \) So, in solution (c), there are 0.00329 moles of sucrose present.